Several years ago I wrote a page with a Java illustration to a solution of problem 4 from the 1995 British Mathematical Olympiad. Earlier today I happened on that page and noticed that the solution refers to point that is not marked in the applet. Since the time I wrote that page, I moved to another computer and to using GeoGebra instead of writing Java applets, so much so that I do not even have my Java development environment on my present computer. To resolve the issue I decided to put together a GeoGebra applet with properly marked and replace the old Java applet. This took probably a couple of minutes. However, this is not what made my day.

At the end of the old page I observed that the configuration in the problem had additional features and pointed to two of them. Now, GeoGebra makes it easy to experiment - form and verify hypothesis, or just plain count on serendipitously stumbling on a dormant feature. It was a sin not to try. The configuration in the problem definitely had many more features waiting to be discovered. Simple though all it was, I am satisfied to have found some.

Problem

has right angle at The internal bisectors of angles and meet and at and respectively. The points and are the feet of the perpendiculars from and to Find angle

The angle was found to be Checking the "Hint" box in the GeoGebra applet below will show the essential steps of the proof.

Checking the "Extra" box will suggested a few more properties: is not the only angle in the diagram that equals (e.g., and some intersections are concyclic; there are several similar triangles (e.g., and

There are probably other properties. Should you find any, please let me know.